TÌM HIỂU VỀ CHUẨN ĐỘ ĐA AXIT - ĐA BAZƠ
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Khi pT = 9 . Tại điểm tương đương thứ hai :<br />
pKa2 + pKa3 pH<br />
II<br />
= = 9,5<br />
2<br />
Ta có: pT = 9 suy ra [H + ]= 10 -9 ,<br />
W 10<br />
−5<br />
=<br />
h<br />
q<br />
II<br />
⎛ W ⎞ C + 2C h − K K<br />
= − ⎜ h − ⎟ −<br />
⎝<br />
h ⎠<br />
2 CC 2( h K h K K<br />
)<br />
2<br />
0 a 2 a3<br />
2<br />
0 +<br />
a 2 +<br />
a 2 a3<br />
0,1 + 2.0,1 (10 ) −10 .10<br />
=− − − =−<br />
q I<br />
−9 2 −7 −12<br />
−9 −5<br />
(10 10 ).<br />
2.0,1.0,1<br />
−9 2 −7 −9 −7<br />
−12<br />
2.((10 ) + 10 .10 + 10 .10<br />
0,0043<br />
0 0<br />
=−0,43
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